Estimate (REF ) is established in [1]}. For the proof of (REF ), we follow the argument of [2]} which is closely related to the analysis in [3]}, [4]}, [5]}, [6]}.
The basic strategy is to perform a perturbative argument with respect to the evolution \(\exp (it\Delta )\)  by a partition on small time intervals which makes the perturbation
\(H_\varepsilon - \Delta \)  better but which losses some regularity on the data because of the summation on the small time intervals.
An additional difficulty resolved in [2]} is coming from the fact that a frequency localisation of \((H_\varepsilon - \Delta )(u)\)  does not imply a frequency localisation of \(u\) .
Let
\({\rm Id}=\sum _{N-{\rm dyadic}}\Delta _N\) 
be a Littlewood-Paley partition of the unity. Therefore the issue is to bound
\(\Vert \Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi \Vert _{L^r( (0,T) ;L^q )}\, .\) 
In order to evaluate (REF ), we distinguish two cases according to the sizes of \(N_1\)  and \(N_2\) 
and to sum up on \(N_1, N_2\) .
First case: \(N_1\ge N_2\)
In this case we split the interval \([0,T]\)  in an essentially disjoint union of intervals of size \( N_1^{-1}\)  as
\([0,T]=\bigcup _j I_j\) 
and we aim to estimate
\(\Vert \Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi \Vert _{L^r(I_j;L^q )}\, .\) 
Suppose that \(I_j=[a,b]\) . Then following [2]} (see also [9]}), for \(t\in [a,b]\)  we can write
\(\Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi =\Delta _{N_1} e^{i(t-a)\Delta } S_{\varepsilon }(a)\Delta _{N_2}\varphi \\+i\int _a^t \Delta _{N_1} e^{i(t-\tau )\Delta } (H_\varepsilon -\Delta ) S_{\varepsilon }(\tau ) \Delta _{N_2}\varphi d\tau .\) 
We now estimate each term in the right hand-side of (REF ). Using [3]}, we estimate the first term as follows for \(\delta >0\) :
\(\Vert \Delta _{N_1} e^{i(t-a)\Delta } S_{\varepsilon }(a)\Delta _{N_2}\varphi \Vert _{L^r(I_j;L^q)}\le CN_1^{-\frac{1}{r}-\delta }\Vert S_{\varepsilon }(a)\Delta _{N_2}\varphi \Vert _{H^{\frac{1}{r}+\delta }}\\\le C(\omega ) |\log \varepsilon |^C N_1^{-\frac{1}{r}-\delta } \Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\) 
where we have used (REF ).
Now we estimate the second term in the right hand-side of (REF ).
Using the Minkowski inequality and [3]}, we can write for every \(\delta >0\) :
\(\Big \Vert \int _a^t \Delta _{N_1} e^{i(t-\tau )\Delta } (H_\varepsilon -\Delta ) S_{\varepsilon }(\tau ) \Delta _{N_2}\varphi d\tau \Big \Vert _{L^r(I_j;L^q)}\\\le C\int _{I_j}\Vert (H_\varepsilon -\Delta ) S_{\varepsilon }(\tau ) \Delta _{N_2}\varphi \Vert _{L^2} d\tau \\\le C(\omega ) |\log \varepsilon |^CN_1^{-1}N_2^{1+ \frac{\delta }{2} }N_2^{-\frac{1}{r}-\delta }\Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\) 
where we have used (REF ) and (REF ).
Summarizing we get
\(\Vert \Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi \Vert _{L^r(I_j;L^q)}\le C(\omega ) |\log \varepsilon |^C \big ( N_1^{-\frac{1}{r}-\delta } +N_1^{-1}N_2^{1-\frac{1}{r}-\frac{\delta }{2}}\big )\Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\) 
and hence using that the number of \(I_j\)  is smaller than \(TN_1\)  taking the \(r\) 'th power of the previous bound and summing on \(j\) , we get the estimate
\(\Vert \Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi \Vert _{L^r((0,T);L^q)}\le C(\omega ) T^\frac{1}{r} |\log \varepsilon |^C \big ( N_1^{-\delta } +N_1^{-1+\frac{1}{r}}N_2^{1-\frac{1}{r}-\frac{\delta }{2}}\big )\Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\) 
and hence
\(\sum _{N_2\le N_1} \Vert \Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi \Vert _{L^r((0,T);L^q)}\le C(\omega ) T^\frac{1}{r}|\log \varepsilon |^C \Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\) 
where we have used
\(\sum _{N_2\le N_1} \big ( N_1^{-\delta } +N_1^{-1+\frac{1}{r}}N_2^{1-\frac{1}{r}-\frac{\delta }{2}}\big )<\infty .\) 
Second case: \(N_1\le N_2\)
We consider again the splitting (REF ) but this time the intervals \(I_j\)  are of size \( N_2^{-1}\) .
Again we consider (REF ) and we estimate each term of the right hand-side.
Since \( N_2^{-1}\le N_1^{-1}\) , using [3]} and (REF ), we estimate the first term at the right hand-side of (REF ) as
\(\Vert \Delta _{N_1} e^{i(t-a)\Delta } S_{\varepsilon }(a)\Delta _{N_2}\varphi \Vert _{L^r(I_j;L^q)}\le C(\omega ) |\log \varepsilon |^CN_2^{-\frac{1}{r}-\delta } \Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\, ,\) 
where \(\delta >0\) .
Next, as above, we can estimate the second term at the right hand-side of (REF ) as
\(\Big \Vert \int _a^t \Delta _{N_1} e^{i(t-\tau )\Delta } (H_\varepsilon -\Delta ) S_{\varepsilon }(\tau ) \Delta _{N_2}\varphi d\tau \Big \Vert _{L^r(I_j;L^q)}\\\le C(\omega ) |\log \varepsilon |^CN_2^{-1}N_2^{1+\frac{\delta }{2}}N_2^{-\frac{1}{r}-\delta }\Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\, .\) 
Summarizing we get
\(\Vert \Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi \Vert _{L^r(I_j;L^q)}\le C(\omega ) |\log \varepsilon |^C \big ( N_2^{-\frac{1}{r}-\delta } +N_2^{-\frac{1}{r}-\frac{\delta }{2}} \big ) \Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\) 
and as above, using that the number of \(I_j\)  is smaller than \(TN_2\)  taking the \(r\) 'th power of the previous bound and summing on \(j\) , we get the estimate
\(\Vert \Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi \Vert _{L^r((0,T);L^q)}\le C(\omega ) T^\frac{1}{r} |\log \varepsilon |^C \big ( N_2^{-\delta } +N_2^{-\frac{\delta }{2}} \big )\Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}.\) 
Hence we get
\(\sum _{N_1\le N_2} \Vert \Delta _{N_1}S_\varepsilon (t)\Delta _{N_2}\varphi \Vert _{L^r((0,T);L^q)}\le C(\omega ) T^\frac{1}{r} |\log \varepsilon |^C \Vert \varphi \Vert _{H^{\frac{1}{r}+\delta }}\) 
since
\(\sum _{N_1\le N_2} \big ( N_2^{-\delta } +N_2^{-\frac{\delta }{2}} \big )<\infty .\) 
We conclude by combining (REF ) and (REF ) with the Minkowski inequality.
